3.3 Points 𝐴(1; 2), 𝐵(5; 1) and 𝐶(3; − 2) are given and shown in the diagram below. Figure 3: Points 𝐴, 𝐵 and 𝐶 on the Cartesian plane. 3.3.1 Determine the coordinates of the point 𝐷 to create parallelogram 𝐴𝐵𝐶𝐷. (3) 3.3.2 Prove that 𝐴𝐵𝐶𝐷 is not a rhombus. (3)

To solve this problem, let's start with the given points \( A(1, 2) \), \( B(5, 1) \), and \( C(3, -2) \) on the Cartesian plane.

### 3.3.1: Determine the coordinates of point \( D \) to create parallelogram \( ABCD \)

In a parallelogram, the diagonals bisect each other. Thus, the midpoint of diagonal \( AC \) should equal the midpoint of diagonal \( BD \).

1. **Calculate the midpoint of \( AC \)**:
   \[
   \text{Midpoint of } AC = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{1 + 3}{2}, \frac{2 - 2}{2} \right) = \left( 2, 0 \right)
   \]

2. **Let the coordinates of \( D \) be \( (x_D, y_D) \)**. We now calculate the midpoint of \( BD \):
   \[
   \text{Midpoint of } BD = \left( \frac{x_B + x_D}{2}, \frac{y_B + y_D}{2} \right) = \left( \frac{5 + x_D}{2}, \frac{1 + y_D}{2} \right)
   \]

3. **Set the midpoints equal**:
   \[
   \left( \frac{5 + x_D}{2}, \frac{1 + y_D}{2} \right) = (2, 0)
   \]

   This gives us two equations:
   \[
   \frac{5 + x_D}{2} = 2 \quad \text{(1)}
   \]
   \[
   \frac{1 + y_D}{2} = 0 \quad \text{(2)}
   \]

4. **Solve for \( x_D \) and \( y_D \)**:
   From equation (1):
   \[
   5 + x_D = 4 \implies x_D = 4 - 5 = -1
   \]

   From equation (2):
   \[
   1 + y_D = 0 \implies y_D = -1
   \]

Thus, the coordinates of point \( D \) are \( D(-1, -1) \).

### 3.3.2: Prove that \( ABCD \) is not a rhombus

To prove that \( ABCD \) is not a rhombus, we need to show that not all sides are of equal length.

1. **Calculate the lengths of the sides**:
   - Length of \( AB \):
   \[
   AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(5 - 1)^2 + (1 - 2)^2} = \sqrt{16 + 1} = \sqrt{17}
   \]

   - Length of \( BC \):
   \[
   BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \sqrt{(3 - 5)^2 + (-2 - 1)^2} = \sqrt{4 + 9} = \sqrt{13}
   \]

   - Length of \( CD \):
   \[
   CD = \sqrt{(x_D - x_C)^2 + (y_D - y_C)^2} = \sqrt{(-1 - 3)^2 + (-1 + 2)^2} = \sqrt{16 + 1} = \sqrt{17}
   \]

   - Length of \( DA \):
   \[
   DA = \sqrt{(x_A - x_D)^2 + (y_A - y_D)^2} = \sqrt{(1 + 1)^2 + (2 + 1)^2} = \sqrt{4 + 9} = \sqrt{13}
   \]

2. **Compare the lengths**:
   - \( AB = \sqrt{17} \)
   - \( BC = \sqrt{13} \)
   - \( CD = \sqrt{17} \)
   - \( DA = \sqrt{13} \)

Since \( AB \neq BC \), the lengths are not all equal, proving that \( ABCD \) is not a rhombus.

### Summary:
- **Coordinates of point \( D \)**: \( D(-1, -1) \)
- **Proof that \( ABCD \) is not a rhombus**: \( AB \) and \( CD \) are equal, but \( BC \) and \( DA \) are equal and not equal to \( AB \).

If you need further details or have questions, let me know! Here are some related questions you might consider:

1. What are the coordinates of the midpoint of diagonal \( BD \)?
2. How do you calculate the area of parallelogram \( ABCD \)?
3. What conditions must be met for a quadrilateral to be a rhombus?
4. Can you derive the slope of each side of the parallelogram?
5. What other types of quadrilaterals can be formed with points \( A \), \( B \), \( C \), and \( D \)? 

**Tip**: Always check the lengths of the sides to determine the properties of quadrilaterals.

3.3 Points A(1; 2), B(5; 1) and C(3; -2) are given and shown in the diagram below. 3.3.1 Determine the coordinates of the point D to create parallelogram ABCD. 3.3.2 Prove that ABCD is not a rhombus.

Explore how to determine the coordinates of point D to form parallelogram ABCD from points A(1, 2), B(5, 1), and C(3, -2). Learn to prove that ABCD is not a rhombus through calculations of side lengths. Suitable for students in Grades 9-10.

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